Question

$\frac{\tan \left(A\right)}{1+\sec \left(A\right)}+\frac{1+\sec \left(A\right)}{\tan \left(A\right)}$ is equal to

• A. $2\sin A$
• B. $2\cos A$
• C. $2\cos ecA$
• D. $2\sec A$

Answer 1

The correct option is C

$2\cos ecA$

Explanation for correct option

Simplifying the expression:

Given, $\frac{\tan \left(A\right)}{1+\sec \left(A\right)}+\frac{1+\sec \left(A\right)}{\tan \left(A\right)}$

On solving,

$$\begin{gathered} \frac{\tan A}{1+\sec A}+\frac{1+\sec A}{\tan A}=\frac{{\tan }^{2}A+{\left(1+\sec A\right)}^2}{\tan A\left(1+\sec A\right)} \\ =\frac{{\sec }^{2}A-1+{\left(1+\sec A\right)}^2}{\tan A\left(1+\sec A\right)}\mathbf{}\left[\because ta{n}^{2}x=se{c}^{2}x-1\right] \\ =\frac{\left(1+\sec A\right)\left(\sec A-1+1+\sec A\right)}{\tan A\left(1+\sec A\right)} \\ =\frac{2\sec A}{\tan A} \\ =\frac{2\cos A}{\cos A\sin A} \\ =2\cos ecA \end{gathered}$$

Hence, the correct answer is Option (C).